Question

我有一份未经编辑的物体清单。这些物体是一 tree树。

物体清单:

new List<Object> 
{
    new { Id = 3, Left = /, Right = / }
    new { Id = 5, Left = /, Right = / }
    new { Id = 4, Left = 2, Right = 5 }
    new { Id = 2, Left = 1, Right = 3 }
    new { Id = 1, Left = /, Right = / }
}

Bin树:

我需要一种算法,能够发现其中任何一点的深度。我知道的唯一算法是中分辨率。这意味着,我必须将物体清单转化为树。认为...... 该网络没有明确的树木数据结构,你如何处理这个问题? 我是否必须把数据结构转换成树(我并不真的想写所有代码)。是否有其他go子?

Answer 1

int nodeToFind = 2;
var currentNode = list.Single(n => n.Id == nodeToFind);
int depth = 0;
while ((currentNode = list
    .FirstOrDefault(i => i.Left == currentNode.Id || i.Right == currentNode.Id)) != null)

    depth++;
Console.WriteLine(depth);

简单但效率低下。

Answer 2

您仅能将物体添加到字典上,将每一左边和正确价值作为钥匙,而Id则作为价值(基本上相反的地图)。接着,你写了你这样的休养职能。

Dictionary<int, int> map;
    int depth(int node)
    {
        if (!map.ContainsKey(node))
            return 0;
        return depth(map[node]) + 1;
    }

Answer 3

You could just make a Dictionary<int,int> parents. Store the node ID as the key, and the node s parent as the value. Note, that this means storing zero, one, or two records in the dictionary for each object in the original list. Then to find the depth of a node, just count the number of times you can get a parent node before you run out. Something like this:

public static int NodeDepth(int node, Dictionary<int,int> parents)
{
     int depth = 0;
     while (parents.ContainsKey(node))
     {
          node = parents[node];
          depth++;
     }
     return depth;
}

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